Question

Number of glucose molecules present in 100 ml 0.1 M aqueous solution of glucose is​

Answer 1

"" ❤️ Answer ❤️ ""

Given, Molarity

=

0.1 M; V=

10 ml=

0.01 litre

Number of moles

=

Molarity×

volume

(in liters)

=

0.1×

0.01

=

10

−3

moles

Number of Molecules

=

moles×

Avogadro number(N

A

)

=

10

−3

×

6.0×

10

23

=

6×

10

20

molecules

So, the correct option is A

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Answer 2

Answer:

$6.022$×$10^{21} molecules$

Explanation:

Given:

Molarity of glucose solution = 0.1 M

Volume = 100 mL

To find: Number of glucose molecules

Explanation:

As Molarity is given, thus from molarity moles of glucose can be calculated. Molarity is defined as the moles of solute dissolved per litre of solution.

$Molarity =\frac{Moles}{volume\ in\ L}$

Thus,

$Moles\ of\ Glucose = Molarity\ of glucose\ solution$×$Volume\ of\ solution\ in\ L$

Moles of Glucose = 0.1 M

Volume of solution in L = 0.100 L

Thus,

$Moles\ of\ glucose =0.1\ M$×$(0.100\ L)$

$Moles\ of\ Glucose= 0.01\ moles$

Next, as it is known that 1 mole of any substance has Avogadro's number of molecules and this number is equal to $6.022$×$10^{23}$.

It means 1 mole of glucose has $6.022$×$10^{23}$ molecules.

And, 0.01 moles of glucose has-

$Number\ of\ molecules\ = (0.01\ moles)$×$(6.022$×$10^{23} molecules/mole )$$= 6.022$×$10^{21}$.

Hence, the number of molecules present in 100 mL of given glucose solution = $6.022$×$10^{21} molecules$

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